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\begin{document}

\title{高等代数一}
\subtitle{13-矩阵乘积的行列式与秩-分块矩阵}
%\institute{上海立信会计金融学院}
\author{{\ppr LQW}}
%\renewcommand{\today}{{\ppr \number\year \,年 \number\month \,月 \number\day \,日} }
\date{{\ppr 2022年11月3日} }

\maketitle

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\begin{frame}{内容提要 }

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\begin{enumerate}
\item  行列式乘积公式
\item  矩阵乘积的秩
\item  分块矩阵的逆阵
\item  分块初等变换

\end{enumerate}

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\begin{frame}{13.1. 可逆矩阵的乘积分解}

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\begin{itemize}
\item  {\color{red}定理：矩阵 $A$ 可逆当且仅当 $A$ 是有限个初等矩阵的乘积。}

\item  例子1：将矩阵 {\footnotesize $A=\begin{pmatrix}1&2 \\ 3&4 \end{pmatrix}$} 写成一些初等矩阵的乘积。

\item  解答：先用行初等变换将矩阵 $A$ 化为单位矩阵，
{\footnotesize 
\begin{eqnarray*}
A = \begin{pmatrix}1&2 \\ 3&4 \end{pmatrix} 
&\xrightarrow[\text{ }]{\text{第1行乘以$-3$加到第2行 }}&
 \begin{pmatrix}1&2 \\ 0&-2 \end{pmatrix} \\
&\xrightarrow[\text{ }]{\text{第2行乘以$-\frac{1}{2}$ }}&
 \begin{pmatrix}1&2 \\ 0&1 \end{pmatrix} \\
&\xrightarrow[\text{ }]{\text{第2行乘以$-2$加到第1行 }}&
 \begin{pmatrix}1&0 \\ 0&1\end{pmatrix} =E_2.
\end{eqnarray*}
}

\end{itemize}

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\begin{itemize}

\item  上述三个行初等变换相当于在矩阵 $A$ 的左边乘以三个相应的初等矩阵，
（根据矩阵乘法的结合律，我们可以不写括号）
{\footnotesize 
\begin{eqnarray*}
\begin{pmatrix}1&-2 \\ 0&1\end{pmatrix}
\cdot 
\begin{pmatrix}1&0 \\ 0&-\frac{1}{2}\end{pmatrix}
\cdot 
\begin{pmatrix}1&0 \\ -3&1\end{pmatrix}
\cdot 
\begin{pmatrix}1&2 \\ 3&4 \end{pmatrix}
= 
\begin{pmatrix}1&0 \\ 0&1\end{pmatrix}.
\end{eqnarray*}
}

\item  用更简略的符号来写，
{\footnotesize 
\begin{eqnarray*}
T_{12}(-2) \cdot D_{2}(-\frac{1}{2}) \cdot T_{21}(-3) \cdot A  =E_2. 
\end{eqnarray*}
}

\item  注意到初等矩阵的逆阵仍然是初等矩阵，将 $A$ 写成初等矩阵的乘积为
{\footnotesize 
\begin{eqnarray*}
A &=& T_{21}(-3) ^{-1} \cdot D_{2}(-\frac{1}{2})^{-1} \cdot T_{12}(-2)^{-1} \\
&=& T_{21}(3) \cdot D_{2}(-2) \cdot T_{12}(2). 
\end{eqnarray*}
}

\end{itemize}

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\begin{frame}{13.3. 小结：矩阵可逆的等价条件 }

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\begin{itemize}
\item  定理：一个 $n$ 阶矩阵 $A$ 是可逆的，当且仅当下述其中之一成立，
\begin{enumerate}
\item  存在 $n$ 阶矩阵 $B$ 使得 $AB=E_n$ 且 $BA=E_n$. （可逆的定义）
\item  存在 $n$ 阶矩阵 $B$ 使得 $AB=E_n$. 
\item  存在 $n$ 阶矩阵 $B$ 使得 $BA=E_n$. 
\item  矩阵 $A$ 的相抵标准形为 $E_n$, 即可以通过初等变换化为单位矩阵。 
\item  矩阵 $A$ 可以写成一些初等矩阵的乘积。
\item  矩阵 $A$ 的秩等于 $n$. 
\item  矩阵 $A$ 的行列式的值不等于零。
\end{enumerate}

\end{itemize}

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\begin{frame}{13.4. 行列式乘积公式 }

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\begin{itemize}
\item  {\color{red}定理：设 $A,B$ 是 $n$ 阶矩阵，则有 $\det(AB)=\det(A)\det(B)$.
}

\item  证明：
\begin{enumerate}
\item  当 $n=1$ 时，一个数的行列式值就是这个数，所以两边自然相等。
\item  当 $n=2$ 时，先计算矩阵的乘积，再计算行列式的值，可得
{\footnotesize 
\begin{eqnarray*}
AB &=& 
\begin{pmatrix}a&b \\ c&d\end{pmatrix} 
\begin{pmatrix}x&y \\ u&v\end{pmatrix}
= \begin{pmatrix}ax+bu&ay+bv \\ cx+du&cy+dv\end{pmatrix} \\ 
\det(AB) &=& (ax+bu)(cy+dv)-(cx+du)(ay+bv).
\end{eqnarray*}
}
先计算行列式的值，再乘积，可得
{\footnotesize 
\begin{eqnarray*}
\det(A)\det(B)=(ad-bc)(xv-uy).
\end{eqnarray*}
}
比较可知上面两个结果相等。
\end{enumerate}

\end{itemize}

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\begin{itemize}
\item  证明（续）：

\begin{enumerate}
\item[3.]   对$n\ge 3$ 的情形，直接验证将非常复杂。在矩阵 $A$ 可逆的情形，我们可以用初等矩阵巧妙地进行证明。

\begin{enumerate}
\item[3.1.]  先对矩阵 $A$ 为初等矩阵的情形进行验证。
当 $A=P_{ij}$, $A=D_i(c)$ 以及 $A=T_{ij}(k)$ 时，有
$%\begin{eqnarray*}
\det(AB)=\det(A)\det(B), 
$%\end{eqnarray*}
这根据行列式的性质可得。

\item[3.2.]  当矩阵 $A$ 为可逆矩阵的情形，将$A$ 写成一些初等矩阵的乘积，例如当 $A=P_2P_1$ 时，
这里 $P_1,P_2$ 是两个初等矩阵，那么有
\begin{eqnarray*}
\det(AB)&=& \det(P_2P_1B) \\
&=& \det(P_2)\det(P_1B) \\
&=& \det(P_2)\det(P_1)\det(B) \\
&=& \det(P_2P_1)\det(B) \\
&=& \det(A)\det(B).
\end{eqnarray*}

\item[3.3.]  当矩阵 $A$ 为不可逆矩阵的情形，考虑矩阵 $A$ 的行最简形。

\end{enumerate}
\end{enumerate}

\end{itemize}

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\begin{frame}{13.6. 乘积矩阵的秩}

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\begin{itemize}

\item  {\color{red}
定理：矩阵乘积的秩，总是不超过每个因子矩阵的秩。
即设 $A,B$ 是两个可以相乘的矩阵，则有
\begin{eqnarray*}
R(AB)&\le& R(A), \\
R(AB)&\le& R(B).
\end{eqnarray*}
}

\vspace{-0.5cm}

\item  证明：一种思路是将矩阵 $A$ 化为相抵标准形。
\begin{enumerate}
\item  存在初等矩阵 $P_1,\cdots, P_s$ 和 $Q_1,\cdots, Q_t$ 使得下式成立，其中 $r=R(A)$,  
{\footnotesize 
\begin{eqnarray*}
P_s\cdots P_1AQ_1\cdots Q_t = \begin{pmatrix} E_r &O \\ O&O \end{pmatrix}=:C. 
\end{eqnarray*}
}

\item  根据初等矩阵与初等变换的关系，以及初等变换不改变矩阵的秩，可得
{\footnotesize 
\begin{eqnarray*}
R(AB) = R(P_1^{-1}\cdots P_s^{-1}CQ_t^{-1}\cdots Q_1^{-1}B) = R(CQ_t^{-1}\cdots Q_1^{-1}B) \le r. 
\end{eqnarray*}
}


\end{enumerate}



\end{itemize}

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\begin{itemize}

\item  {\color{red}
定理：一个可逆矩阵乘以另一个矩阵，不改变另一个矩阵的秩。
即设 $A$ 是可逆矩阵，设 $B$ 是一个与 $A$ 可以相乘的矩阵，则有
\begin{eqnarray*}
R(AB)= R(B).
\end{eqnarray*}
}

\item  证明：
\begin{enumerate}
\item  因为矩阵 $A$ 可逆所以它是一些初等矩阵 $P_s,\cdots,P_2,P_1$ 的乘积。
\item  乘以一个初等矩阵等价于做相应的初等变换。
\item  初等变换不改变矩阵的秩。
\end{enumerate}
\begin{eqnarray*}
R(AB)&=& R(P_s\cdots P_2P_1B) 
=\cdots = R(P_2P_1B) 
= R(P_1B) 
= R(B). 
\end{eqnarray*}

\end{itemize}

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\begin{itemize}

\item  例子2：设矩阵 $A,B$ 及其分块如下，计算乘积$AB$, 
{\footnotesize 
\begin{eqnarray*}
A=\begin{pmatrix} 1&0&0&0 \\ 0&1&0&0 \\ -1&2&1&0 \\ 1&1&0&1   \end{pmatrix}
=\begin{pmatrix} E_2&O \\ A_1&E_2  \end{pmatrix}, 
\hspace{0.2cm}
B=\begin{pmatrix} 1&0&3&2 \\ -1&2&0&1 \\ 1&0&4&1 \\ -1&-1&2&0   \end{pmatrix}
=\begin{pmatrix} B_1&B_2 \\ B_3&B_4   \end{pmatrix}. 
\end{eqnarray*}
}

\item  解答：按分块矩阵的乘法规则，可得
{\footnotesize 
\begin{eqnarray*}
AB=\begin{pmatrix} E_2&O \\ A_1&E_2  \end{pmatrix}
\begin{pmatrix} B_1&B_2 \\ B_3&B_4   \end{pmatrix}
=\begin{pmatrix} B_1 & B_2 \\ A_1B_1+B_3 & A_1B_2+B_4   \end{pmatrix}. 
\end{eqnarray*}
}

\end{itemize}

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\begin{itemize}

\item  例子3：设 $A,B$ 是 $n$ 阶可逆矩阵。求矩阵
{\footnotesize 
$%\begin{eqnarray*}
P=\begin{pmatrix} A&C \\ O&B  \end{pmatrix}
$%\end{eqnarray*}
}
的逆阵。

\item  解答：设逆阵为 {\footnotesize $ X=\begin{pmatrix} X_1&X_2 \\ X_3&X_4  \end{pmatrix} $}. 
根据逆阵的定义，有 $PX=E$, 即 
{\footnotesize 
\begin{eqnarray*}
PX=
\begin{pmatrix} A&C \\ O&B  \end{pmatrix}
\begin{pmatrix} X_1&X_2 \\ X_3&X_4  \end{pmatrix}
=
\begin{pmatrix} E&O \\ O&E  \end{pmatrix}. 
\end{eqnarray*}
}

于是可得下述四个等式，从中可以求出 $X_1, X_2, X_3, X_4$. 
{\footnotesize 
\begin{eqnarray*}
\left\{
\begin{array}{rcl}
AX_1 + CX_3 &=& E, \\
AX_2 + CX_4 &=& O, \\ 
BX_3 &=& O, \\ 
BX_4 &=& E. 
\end{array}\right. 
\end{eqnarray*}
}

\end{itemize}

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\begin{itemize}

\item  例子4：设 $A_1,A_2, B_1, B_2$ 都是 $n$ 阶矩阵。
设矩阵 {\footnotesize $A=\begin{pmatrix} A_1&O \\ O&A_2  \end{pmatrix}$}, 
矩阵 {\footnotesize $B=\begin{pmatrix} B_1&O \\ O&B_2  \end{pmatrix}$}.  
求它们的和 $A+B$, 乘积 $AB$ 与逆阵 $A^{-1}$. 

\item  解答：按照分块矩阵的运算规则，可得 
{\footnotesize 
\begin{eqnarray*}
A+B &=& \begin{pmatrix} A_1&O \\ O&A_2  \end{pmatrix} + \begin{pmatrix} B_1&O \\ O&B_2  \end{pmatrix}
= \begin{pmatrix} A_1+B_1&O \\ O&A_2+B_2  \end{pmatrix}, \\ 
AB &=& \begin{pmatrix} A_1&O \\ O&A_2  \end{pmatrix}  \begin{pmatrix} B_1&O \\ O&B_2  \end{pmatrix}
= \begin{pmatrix} A_1B_1&O \\ O&A_2B_2  \end{pmatrix}, \\ 
A^{-1} &=& \begin{pmatrix} A_1&O \\ O&A_2  \end{pmatrix} ^{-1} 
= \begin{pmatrix} A_1^{-1}&O \\ O&A_2^{-1}  \end{pmatrix}. 
\end{eqnarray*}
}

\end{itemize}

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\begin{enumerate}

\item  例子5：设 $A,B,C,D,K$ 都是 $n$ 阶矩阵。设 $2n$ 阶矩阵 $M,S,T$ 如下，
{\footnotesize 
\begin{eqnarray*}
M=\begin{pmatrix} A&B \\ C&D  \end{pmatrix}, 
\hspace{0.2cm} 
S=\begin{pmatrix} E&O \\ K&E  \end{pmatrix}, 
\hspace{0.2cm} 
T=\begin{pmatrix} E&K \\ O&E  \end{pmatrix}. 
\end{eqnarray*}
}
计算分块矩阵的乘积 $SM, MS$ 与 $TM, MT$. 

\item  解答：分块行初等变换相当于左乘分块初等矩阵，
{\footnotesize 
\begin{eqnarray*}
SM &=& \begin{pmatrix} E&O \\ K&E  \end{pmatrix}\begin{pmatrix} A&B \\ C&D  \end{pmatrix}
= \begin{pmatrix} A&B \\ KA+C&KB+D  \end{pmatrix}, \\ 
TM &=& \begin{pmatrix} E&K \\ O&E  \end{pmatrix}\begin{pmatrix} A&B \\ C&D  \end{pmatrix}
= \begin{pmatrix} A+KC&B+KD \\ C&D  \end{pmatrix}. 
\end{eqnarray*}
}

\end{enumerate}

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\begin{enumerate}

\item  例子6：设有如下分块矩阵，
{\footnotesize 
\begin{eqnarray*}
S=\begin{pmatrix} E&O \\ K&E  \end{pmatrix} 
= \begin{pmatrix} 1&0&0&0 \\ 0&1&0&0 \\ {\color{red}k_1}&{\color{red}k_2}&1&0 \\ {\color{red}k_3}&{\color{red}k_4}&0&1  \end{pmatrix},  
\hspace{0.2cm} 
M=\begin{pmatrix} A&B \\ C&D  \end{pmatrix} 
= \begin{pmatrix} a_1&a_2&b_1&b_2 \\ a_3&a_4&b_3&b_4 \\ c_1&c_2&d_1&d_2 \\ c_3&c_4&d_3&d_4  \end{pmatrix}.  
\end{eqnarray*}
}
计算 $SM$, 并验证一次分块初等变换相当于若干次真正的初等变换。

\item  解答：计算可知，对矩阵 $M$ 进行四次第三类行初等变换可以得到 $SM$. 另一方面，矩阵 $S$ 是四个初等矩阵的乘积，即有 
{\footnotesize 
\begin{eqnarray*}
S = T_{31}({\color{red}k_1})T_{41}({\color{red}k_3})T_{32}({\color{red}k_2})T_{42}({\color{red}k_4}). 
\end{eqnarray*}
}


\end{enumerate}

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\begin{enumerate}

\item  设有三阶矩阵 {\footnotesize $A=\begin{pmatrix} a_1&a_2&a_3 \\ b_1&b_2&b_3 \\ c_1&c_2&c_3 \end{pmatrix}$}. 
设 $R(A)=1$. 证明存在列向量 {\footnotesize $\alpha=\begin{pmatrix} u_1 \\ u_2 \\ u_3  \end{pmatrix}$} 
与列向量 {\footnotesize $\beta=\begin{pmatrix}  x_1 \\ x_2 \\ x_3 \end{pmatrix}$} 使得 $A=\alpha\beta^t$, 即 
{\footnotesize 
$\begin{pmatrix} a_1&a_2&a_3 \\ b_1&b_2&b_3 \\ c_1&c_2&c_3 \end{pmatrix}
= \begin{pmatrix} u_1 \\ u_2 \\ u_3  \end{pmatrix}
\begin{pmatrix}  x_1 & x_2 & x_3 \end{pmatrix}. 
$
}

\item  设矩阵 {\footnotesize $A=\begin{pmatrix} 2&-1&0&0 \\ -3&2&0&0 \\ 31&-19&3&-4 \\ -23&14&-2&3 \end{pmatrix}$}. 用分块矩阵的方法求逆阵 $A^{-1}$. 

\end{enumerate}

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\begin{enumerate}

\item  解答思路：对矩阵 $A$ 做行初等变换，根据 $R(A)=1$, 可得行最简形只有一行不全为零。

\item  解答思路：分块之后成为分块下三角矩阵。仿照例子3求逆，可得
{\footnotesize 
\begin{eqnarray*}
A^{-1}=\begin{pmatrix}  2&1&0&0 \\ 3&2&0&0 \\ 1&1&3&4 \\ 2&-1&2&3  \end{pmatrix}. 
\end{eqnarray*}
}

%A: 你的热爱是什么？
%B: 写ppt. 

\end{enumerate}

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